Describing behavior on each side of a vertical asymptote

Jacobpm64
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Find the vertical asymptotes of the graph of F(x) = (3 - x) / (x^2 - 16)

ok if i factor the denominator.. i find the vertical asymptotes to be x = 4, x = -4.

The 2nd part of the problem asks:
Describe the behavior of f(x) to the left and right of each vertical asymptote.. I'm not sure what i need to write for this.
 
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It's asking for what happens as the function appraches the asymptotes from the left and the right, does it go to infinity - what?

You can use limits to find out.
 
ahh thanks.. so...

f(x) approaches +inf as it approaches x = -4 from the left...
f(x) approaches -inf as it approaches x = -4 from the right...
f(x) approaches +inf as it approaches x = 4 from the left...
f(x) approaches -inf as it approaches x = 4 from the right...

correct?
 
Correct. :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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